solex2 - Solutions to the second midterm exam (November 14)...

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Unformatted text preview: Solutions to the second midterm exam (November 14) Problem 1. Matrix A defines a linear transformation R4 −→ R3 , x −→ Ax. Hence the kernel of A lies in R4 and the image of A lies in R3 . Bringing A to the reduces row echelon form, we obtain 113 2 1 5 317 4 11 5 −→ 0 −1 6 0 −2 3 −1 −2 4 11 −3 −→ 0 1 −6 00 34 1 1 3 −→ 0 00 0 021 1 1 3. 000 Hence the kernel of A consists of the vectors −1 −2 −2x3 − x4 −3 −1 −x3 − 3x4 , + x4 = x3 0 1 x3 1 0 x4 −1 −2 −1 −3 where x3 and x4 can be any numbers. We conclude that is a basis of , 0 1 1 0 the kernel A. Selecting the columns of A corresponding to the leading 1’s, we conclude of 1 1 2 , 1 is a basis of the image of A. that 3 1 −1 −2 −1 −3 , Answer: Vectors form a basis of the kernel of A. Geometrically, the 0 1 1 0 1 1 4 2 , 1 form a basis of the image of A. kernel of A is a plane in R . Vectors 3 1 Geometrically, the image of A is a plane in R3 . Problem 2. The cosine of the angle between a + b and a − b is a+b · a−b . a−b a+b Now, a+b · a−b =a·a+b·a−a·b−b·b = a 1 2 2 −b = 4 − 9 = −5, a+b = = a+b · a+b = a 2 a·a+b·a+a·b+b·b 2 + 2a · b + b = √ 9+4+4 = √ 17 and a−b = = a−b · a−b = a 2 a·a−b·a−a·b+b·b 2 − 2a · b + b = √ 9 − 4 + 4 = 3. √ Hence the cosine of the angle is −5/3 17. 5 Answer: The cosine of the angle between a + b and a − b is − √ . 3 17 Problem 3. Matrix A defines a linear transformation R3 −→ R5 , x −→ Ax, while matrix B defines a linear transformation R4 −→ R3 , x −→ Bx. a) Since dim ker A + dim im A = 3 and dim im A = rank A = 2, we conclude that dim ker A = 1. Answer: the dimension of the kernel of A is 1. b) Since dim ker B + dim im B = 4 and dim im B = rank B = 2, we conclude that dim ker B = 2. Answer: the dimension of the kernel of B is 2. c) We have (AB )x = A (Bx) for all x in R4 . The dimension of the image of B is 2, so, geometrically the image of B is a plane in R3 . Now, the kernel of A is a line. If the kernel of A lies in the image of B then the dimension of the image of AB will be 1, so the dimension of the kernel of AB will be 4 − 1 = 3. If the kernel of A intersects the image of B in the origin only, the dimension of the image of AB will be 2, so the dimension of the kernel of AB will be 4 − 2 = 2. ker (A) im( B ) im(AB) ker (A) im( B ) 2 im(AB) In fact, both values can occur, for example, when 1 0 A = 0 0 0 0 1 0 0 0 and the dimension 10 0 1 A = 0 0 00 00 0 0 0 0 0 and 10 B = 0 1 00 of the kernel of AB is 0 0 0 0 and B = 1 0 0 0 00 0 0 00 then 1 0 AB = 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 then 0 1 AB = 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2. If 0 0 1 00 0 0 00 and the dimension of the kernel of AB is 3. Answer: the possible dimensions of the kernel of AB are 2 and 3. Problem 4. Writing the system in the form Ax = b, where 1 −1 0 2 −1 , b = 1 and x = x , A= y 31 3 We solve the system AT Ax = AT b, where AT A = AT b = 1 −1 2 −1 1 2 −1 −1 3 1 3 1 1 −1 2 −1 = 14 0 03 31 0 1 = 11 , 2 3 and that is, 14x = 11 and 3y = 2, from which x = 11/14 and y = 2/3. Answer: the least squares solution is x = 11/14 and y = 2/3. Problem 5. a) We look for the projection w in the form w = xa + y b for some x and y such that (v − w) · a = 0 and (v − w) · b = 0. That is, 1 1 0 1 0 − x 1 − y 1 · 1 = 0 1 0 1 0 and 3 1 1 0 0 0 − x 1 − y 1 · 1 = 0. 1 0 1 1 Computing the dot products, we obtain 2x + y = 1 and x + 2y = 1, from which x = y = 1/3 and 1 0 1/3 1 1 w = 1 + 1 = 2/3 . 3 3 0 1 1/3 1/3 Answer: the orthogonal projection is 2/3 . 1/3 b) Let u be the orthogonal projection of v onto the orthogonal complement to the plane spanned by a and b. u v w Then u + w = v , so 1 1/3 2/3 u = v − w = 0 − 2/3 = −2/3 . 1 1/3 2/3 2/3 Answer: the orthogonal projection is −2/3 . 2/3 4 ...
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This note was uploaded on 12/26/2011 for the course MATH 214 taught by Professor Conger during the Fall '08 term at University of Michigan.

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